Chapter 1, New Ways to Think about Bits

2. 
3. 
4. 
5. 
6. 
7. 

8. import numpy as np
   A = np.matrix( [[1, 2, 3, 0],
                   [2, 1, -1, 3]] )
   B = np.matrix( [[1, 1, -2],
                   [3, 2, -1],
                   [0, 4, 3],
                   [3, -3, 5]] )
   print(np.dot(A, B))

import numpy as npA = np.matrix( [[2],
                [3],
                [1]] )
B = np.matrix( [[8, 4, 0],
            ...

Chapter 2, What Is a Qubit?

1. 1,000.
2. 0.
3. 500.
4. Approximately 250.
5. The number of zeros is usually between 490 and 510. Very rarely does the number become lower than 480 or higher than 520.

6. from qiskit import QuantumCircuit
   circuit = QuantumCircuit(3, 4)
   circuit.h(0)
   circuit.h(2)
   circuit.barrier()
   circuit.measure([1], [2])
   circuit.measure([2], [1])
   circuit.measure([0], [3])
   display(circuit.draw('latex'))

Chapter 3, Math for Qubits and Quantum Gates

1. A. represents a qubit because .

B. doesn’t represent a qubit because .

C. doesn’t represent a qubit because it has three entries, not two.

2. Here is the result:

3. 
4. 
5. Here is the Qiskit code:

   qc = QuantumCircuit(1)
   qc.h(0)
   qc.z(0)

6. 
7. According to Figure 3.25, . So, we have .
   
8. Here is the matrix calculation:



9. Here’s the Hadamard gate acting on |+⟩:


Here’s the gate acting on |+⟩:


10. 

Chapter 4, Qubit Conspiracy Theories

1. q1 is 1, q0 is 0.
2. You can demonstrate that writing in the form is impossible. Here’s how:

Assume (to the contrary) that . Then, .

But since , either or .

In the first case, must be 0 instead of 1. In the second case, must be 0 instead of 1. Either way, things don’t work out.



4. from qiskit import QuantumCircuit
   circ = QuantumCircuit(2)
   circ.h(0)
   circ.cnot(0, 1)
   circ.x(1)
   circ.z(1)
   display(circ.draw('latex', initial_state=True))

The resulting sphere looks like this:



The |01⟩ qubit’s phase is , and the |10⟩ qubit’s phase is 0.

5. 
6. There are eight ways to assign up-or-down arrows to the three measuring directions on the left. For the six ways that resemble the scenario in Figure 4.35, the probability of disagreement is . For the remaining two ways, the probability of disagreement is 1 (see...

Chapter 5, A Fanciful Tale about Cryptography

1. Send information about any errors that you find to [e-mail].
2. Send information about any errors that you find to [e-mail].
3. If Eve measures Alice’s qubits, then Alice and Bob will know about it. So, Alice and Bob won’t use any of those qubits as their encryption key. So, if Eve knows some of Alice’s zeros and ones, that information won’t be useful to her.
4. The following example illustrates the use of the right-distributive law for tensor products:

5. When we cross-multiply, we use both the left- and right-distributive laws. In the following equations, one step uses the right-distributive law:
   

And the next step uses the left-distributive law:



We use the scalar multiplication law when we combine two qubits’ scalars. Here’s an example:


6. You can code the circuit...

Chapter 6, Quantum Networking and Teleportation

1. In the Quantum operations for teleportation section, we derive the following formula:


When the initial state of Alice’s qubit is |0⟩, = 1 and = 0. So, the formula becomes



We have four possibilities, and each possibility has three qubits. Alice measures the middle qubit. When that measurement yields the value 1, Bob applies the X gate. So, we have



No matter what values Alice’s measurements yield, Bob’s qubit is in the |0⟩ state. So the value of Alice’s qubit has been teleported to Bob.

The calculation is similar when the initial state of Alice’s qubit is |1⟩.

2. When you hardcode alpha = 0.8228 and beta = 0.5683, a call to add_gates gives you an error message. The message is Sum of amplitudes-squared does not equal one. This happens because the value of |0.8228|2 + |0.5683|2 isn’...

Chapter 7, Deutsch’s Algorithm

1. This circuit is not reversible. When x = y, the output is 01 regardless of whether x and y are both 0 or x and y are both 1. When x y, the output is 10 regardless of which input is 0 and which is 1.
2. The number of correct shots varies quite a lot because the amount of noise isn’t predictable. Also, some quantum computers tend to be less noisy than others. On IBM devices, the number of correct shots out of 100 is typically in the 90s but sometimes in the 80s.
3. The CNOT gate affects both the top and bottom qubits. But when we add an X gate, we add it to the bottom qubit. The top qubit (the qubit that we eventually measure) remains unchanged.
4. This circuit implements the Opposite_of function. The leftmost X gate reverses the roles of inputs 0 and 1 from what they’d be with the Identity function. Then, the rightmost X gate restores the top qubit to its original 0 value or 1 value. That rightmost X gate has no effect on...

Chapter 8, Grover’s Algorithm

1. It takes 20 qubits to search among 1,000,000 values because 220 = 1,000,000. The number of Grover iterations will be .

3. Here’s the code:

   import random
   oracle_matrix = [
       [1, 0,  0, 0, 0, 0, 0, 0],
       [0, 1,  0, 0, 0, 0, 0, 0],
       [0, 0,  1, 0, 0, 0, 0, 0],
       [0, 0,  0, 1, 0, 0, 0, 0],
       [0, 0,  0, 0, 1, 0, 0, 0],
       [0, 0,  0, 0, 0, 1, 0, 0],
       [0, 0,  0, 0, 0, 0, 1, 0],
       [0, 0,  0, 0, 0, 0, 0, 1]
   ]
   entry = random.randint(0, 7)
   print(entry)
   oracle_matrix[entry][entry] = -1
   oracle = QuantumCircuit(3)
   oracle.unitary(oracle_matrix, qubits=[0, 1, 2], label=’oracle’)
   oracle.barrier()
   display(oracle.draw('latex'))

4. The expression...

Chapter 9, Shor’s Algorithm

1. We start by finding values of 3n % 14:


Next, we calculate the following:



When we calculate 14/13, we find that it isn’t an integer. But 14/2 = 7. So, 14 = 2·7.

2. We start by finding values of 2n % 35:


Next, we calculate the following:



When we calculate 35/13 and 35/3, we find that these aren’t integers. But 35/5 = 7. So, 35 = 5·7.

3. Dividing 71 by 8 gives us 8 with a remainder of 7. So, is the same as . According to Figures 9.9 and 9.13, is equal to .
4. The QFT and QFT† matrices are shown here:

5. The 2 × 2 QFT matrix is the Hadamard matrix because the second roots of unity are 1 and -1.
6. The missing values are shown here:
   
7. In the coprime powers sequence for 22 (with coprime 3), the period is 5. But 5 isn’t divisible by 2. You can’t find 35/2 + 1 or 35/2 &...